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Illusion
1 month ago
8

You are camping in the wilderness. After a few days, you are horrified to discover that you did not pack as many batteries as yo

u had planned, and you have no working batteries for your lights at night. Rummaging through the spare parts in the back of your truck, you find an old motor. On the plate, the information claims that the motor operates from 120 V, rotating at 1,600 rev/min, with an average back emf of 80.0 V. You wish to use the motor as a generator to provide a voltage with a peak value of 7.50 V to operate your electric lantern. You attach a hand crank to the armature of the motor. You need to determine the angular speed (in rev/s) at which you must rotate the crank to provide the desired voltage. Model the armature as a flat coil of wire. Notice that the average back emf is provided, not the peak value, so you will need to find an expression for the average back emf of a motor in terms of parameters associated with the armature rev/s Need Help?
Physics
1 answer:
Keith_Richards [3.2K]1 month ago
8 0
Angular speed is calculated as 2.5 rev/sec.
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You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a
serg [3582]

Answer:

The accurate statements are

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant velocity in a circular path.

8. The train must be undergoing acceleration.

Explanation:

As we observe that the string forms an angle with the horizontal

we can formulate the force equation relevant to the given ball

F_x = Tcos\theta

ma = Tcos45

similarly in the Y direction

mg = Tsin45

Thus we conclude

\frac{ma}{mg} = cot 45

a = g cot45

This leads us to deduce that the train is accelerating with an acceleration identical to that of gravity

The correct statements will be

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant speed in a circular path.

8. The train must be experiencing acceleration.

8 0
1 month ago
Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a
Sav [3153]

Reminder: The illustration mentioned in the inquiry is included here as a file.

Response:

The intensity of the magnetic field is B = \frac{0.071 \mu I}{d}

Clarification:

The magnetic field can be calculated using the following equation:

B = \frac{\mu I}{2\pi R}...............(1)

Where I refers to the current flowing through the wires

To find the distance R from point 1 to m, apply the Pythagorean theorem

R = \sqrt{d^{2} + (2d)^{2} }

R = \sqrt{5d^{2} } \\R = d\sqrt{5}

Inserting R into equation (1)

B = \frac{\mu I}{2\pi d\sqrt{5} }

B = \frac{0.071 \mu I}{d}

3 0
1 month ago
Two carts are involved in an elastic collision. Cart A with mass 0.550 kg is moving towards Cart B with mass 0.550 kg, which is
Sav [3153]

Answer:

A) v_b=0.8\ m.s^{-1} denotes the resultant velocity of cart B post-collision.

B) KE_A=0.176\ J

C) KE_B=0\ J

D) ke_a=0\ J

E) ke_B=0.176\ J

F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.

G) Yes, momentum is conserved in every elastic collision.

Explanation:

Given:

  • mass of car A, m_a=0.55\ kg
  • mass of car B, m_b=0.55\ kg
  • initial velocity of car A, u_a=0.8\ m.s^{-1}
  • final velocity of car A, v_a=0\ m.s^{-1}

A)

The question mentions the cars experience an elastic collision:

By applying momentum conservation principles:

m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b

0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b

v_b=0.8\ m.s^{-1} denotes the resulting velocity of cart B after collision.

B)

Initial kinetic energy of cart A:

KE_A=\frac{1}{2} m_a.u_a^2

KE_A=0.5\times 0.55\times 0.8^2

KE_A=0.176\ J

C)

Initial kinetic energy of cart A:

KE_B=\frac{1}{2} \times m_b.u_b^2

KE_B=0.5\times 0.55\times 0^2

KE_B=0\ J

D)

The final kinetic energy of cart A:

ke_A=\frac{1}{2} m_a.v_a^2

ke_a=0.5\times 0.55\times 0^2

ke_a=0\ J

E)

The final kinetic energy of cart B:

ke_B=\frac{1}{2} m_b.v_b^2

ke_B=0.5\times 0.55\times 0.8^2

ke_B=0.176\ J

F)

Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.

G)

Indeed, momentum is consistently conserved in elastic collisions.

5 0
1 month ago
what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
Sav [3153]
Recall this formula for a device operating in a direct current circuit:
P = IV
In this equation, P stands for the power emitted by the device, I signifies the current passing through the device, and V represents the voltage drop across it.

Using ampere for current and volt for voltage means that multiplying current by voltage gives you power measured in watts.
5 0
2 months ago
If the position of an object is zero at one instant, what is true about the velocity of that object?
ValentinkaMS [3465]
If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.
6 0
2 months ago
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